The 11 Most Beautiful Mathematical Equations

in proofs and equations

rare - but, oh, so satisfying

1+e^i*π = 0

I hope that pi shows up correctly

What makes it kind of a kludge is what it's actually describing, i.e. that one π is only one half a turn so you need to add 1 to get back to zero:

[center]
e [sup]iπ[/sup] + 1 = 0[/center]

e [sup]iπ[/sup] is the curved arrow leading to -1 on the horizontal axis.
+1 is the vector leading you from -1 back to 0.

But using a circle constant based upon radius, τ, one can use Euler to now describe one complete turn of the circle returning back to 1.

[center]
e [sup]iτ[/sup] = 1[/center]

1 tau is 1 turn, not half a turn. Easy and intuitive. Math should illuminate, not obfuscate.

http://tauday.com/

. . . it contains the zero, which is itself an important mathematical concept.

[center]

e [sup]iτ[/sup] - 1 = 0[/center]

e[sup]iτ[/sup] is the curved arrow making one complete circuit leading back to +1
-1 is the vector taking you from +1 to 0.

Math is also mystery, and some of the mystery of real part of zeta function being exactly between 0 and 1, according to Riemann (but so far unproven), seems to have something to do with Euler's Identity...

e^(i*x) = cos x + i sin x

is even better, since it provides a way to compute many trig identities, such as

sin^2 x + cos^2 x = (cos x + i sin x)(cos x - i sin x) = e^(i*x) * e^(-i*x) = 1

or

cos 2x = (e^(2i*x) + e^(-2i*x))/2 = (e^(i*x)^2 + e^(-i*x)^2)/2 = <-2 + (e^(i*x) + e^(-i*x))^2>/2
= -1 + 2cos^2 x = -sin^2 x - cos^2 x + 2 cos^2 x = cos^2 x - sin^2 x

along with the Fourier Transform...

Pythagorean Theorem:

[center][/center]
Pythagorean theorem was listed in the OP article but that's one nifty geometric proof.

Maxwell's equations (classical electrodynamics):

[center]


[/center]
Once described all we knew about electromagnetics and it's still good enough for government work.

Quantum electrodynamics:

[center][/center]
Shout out to Richard Feynman!

Leibniz formula for Pi:

[center]π/4 = 1/1 - 1/3 + 1/5 - 1/7 + 1/9 - 1/11 + 1/13 - ...[/center]
It's about the slowest converging series ever but it gets bonus points for its simplicity.

Quadratic factorization:

[center][/center]
I memorized it in grade school and never forgot it.

And then there's this rather surprising one...

[center]e^π - π = 20[/center]
Commonly used to test the accuracy of floating point operations on calculators.

that the so-called Gaussian integers a + bi have unique prime factorization, and this unlocks many facts about Pythagorean triples, because |a + bi|^2 = a^2 + b^2

For example, which ordinary primes appear as the hypotenuse of a right triangle? Exactly the ordinary primes that do not remain prime when considered as Gaussian integers: that is, the prime 2 and the primes of the form 4n + 1

For example, if (a,b,c) is a primitive Pythagorean triple, c has no factor of the form 4n + 3. For, consider the equation a^2 + b^2 = (a + bi)(a - bi) = c^2. If we completely factor a + bi in the Gaussian integers and take the complex conjugate of each factor, we get the factorization of a - bi. The prime factorizations of a^2 + b^2 and of c^2 must be the same. If c has an ordinary prime factor q of the form 4n + 3, q does not factor further, so q appears among the factors of a + ib; then q divides each of a, b, c -- contradicting the fact that (a,b,c) is primitive

And so on ...

Again, if (a,b,c) is a primitive Pythagorean triple, consider the (ordinary) prime factorization of c: c = p1*p2*...*pn. No ordinary prime pj here has the form 4n + 3, so each of these ordinary primes factors in the Gaussian integers: pj = (uj + i*vj)(uj - i*vj). Multiplying
(u1 + i*v1)(u2 + i*v2)...(un + i*vn) = (u + i*v)
we find c = (u + i*v)(u - i*v) = u^2 + v^2. So the hypothenuse of a primitive Pythagorean triple is itself a sum of two squares: that, is, a primitive Pythagorean triple is of the form (a,b,u^2 + b^2) and the Pythagorean theorem in this case looks like
a^2 + b^2 = (u^2 + v^2)^2

Using ordinary arithmetic, it is clear that if (a,b,c) is a primitive Pythagorean triple, then a and b cannot both be even; and if they were both odd a^2 + b^2 would be even (but not divisible by 4), so c^2 would be even (but not divisible by 4), so c would be even, so c^2 would be divisible by 4, a contradiction. Therefore one of the legs is even and the other odd. Interchanging a and b if necessary, we may assume a is odd and b is even. Therefore c = u^2 + v^2 = (u + i*v)(u - i*v) is also odd. Since c is odd, the Gaussian prime 1 + i divides neither (u + i*v) nor (u - i*v). Moreover
(a + ib)(a - ib) = a^2 + b^2 = c^2 = (u^2 + v^2)^2
so the Gaussian prime 1 + i divides neither (a + ib) nor (a - ib). But any Gaussian prime that divides both (a + ib) and (a - ib) must be either an ordinary prime of the form 4n + 3 or the Gaussian prime 1 + i, and so (a + ib) and (a - ib) are relatively prime. Each distinct prime factor of c^2 must therefore appear uniquely in (a + ib) and (a - ib); such a prime factor cannot appear in both. So
(a + ib)(a - ib) = c^2 implies both (a + ib) and (a - ib) are squares, say, a + ib = (x + iy)^2 and a - ib = (x - iy)^2
Writing out
a + ib = (x + iy)^2 = x^2 - y^2 + i*2xy
we have a = x^2 - y^2 and b = 2xy
while
(x^2 + y^2)^2 = (x + iy)^2(x - iy)^2 = (a + ib)(a - ib) = a^2 + b^2 = c^2
gives c = x^2 + y^2
So we recover the ancient Greek formula for primitive Pythagorean triples: they all have the form
(x^2 - y^2,2xy,x^2 + y^2)
where exactly one of x and y is even

Even more astonishing is just how very close to an integer exp(pi*sqrt(163)) is

(but it is much more challenging to compute)

sorry



The iteration x <- ln(x + 20) converges pretty quickly to the solution of e^x - x = 20, giving

x = 3.1416333028

http://en.wikipedia.org/wiki/Mathematical_coincidence

Some fun ones...

2[sup](1/12)[/sup] * 5[sup](1/7)[/sup] = 1.33333319249

π [sup]3[sup]2[/sup][/sup]/e[sup]2[sup]3[/sup][/sup] = 9.9998387978